QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#572743 | #9319. Bull Farm | KeeperHihi | TL | 93ms | 24700kb | C++20 | 3.5kb | 2024-09-18 16:12:19 | 2024-09-18 16:12:19 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
struct Info {
int cur = 0;
int mx = -1e9;
friend bool operator < (Info a, Info b) {
return a.mx > b.mx;
}
};
int cal(char x, char y) {
return (x - 48) * 50 + y - 48;
}
void solve() {
int n, l, q;
cin >> n >> l >> q;
vector to(l + 1, vector<int>(n + 1));
vector inv(l + 1, vector<int>(n + 1));
vector<vector<pair<int, int>>> adj(n + 1);
for (int i = 1; i <= l; i++) {
string s;
cin >> s;
vector<int> vis(n + 1);
for (int j = 0, k = 1; j < s.size(); j += 2, k++) {
to[i][k] = cal(s[j], s[j + 1]);
inv[i][to[i][k]] = k;
vis[to[i][k]] = 1;
}
int cnt = accumulate(vis.begin() + 1, vis.end(), 0);
if (cnt == n) {
for (int j = 1; j <= n; j++) {
adj[j].emplace_back(to[i][j], i);
adj[to[i][j]].emplace_back(j, i);
}
} else if (cnt == n - 1) {
int x = 0;
vis.assign(n + 1, 0);
for (int j = 1; j <= n; j++) {
if (vis[to[i][j]]) {
x = to[i][j];
break;
}
vis[to[i][j]] = 1;
}
vector<int> a;
for (int j = 1; j <= n; j++) {
if (to[i][j] == x) {
a.emplace_back(j);
}
}
assert(a.size() == 2);
// 初始只能在 a[0] 或者 a[1],其结果都是唯一确定的,把对应的几个点 merge 一下就行了
auto dfs = [&](auto self, int u, int fa, int t) -> void {
if (vis[u]) {
// cout << "edge = " << t << ' ' << fa << endl;
adj[t].emplace_back(fa, i);
return;
}
vis[u] = 1;
self(self, inv[i][u], u, t);
};
vis.assign(n + 1, 0);
vis[0] = 1;
inv[i][x] = a[1];
dfs(dfs, a[0], a[0], a[0]);
vis.assign(n + 1, 0);
vis[0] = 1;
inv[i][x] = a[0];
dfs(dfs, a[1], a[1], a[1]);
// 注意是单向边!!!
}
}
vector<vector<tuple<int, int, int>>> qry(n + 1);
vector<int> ans(q);
for (int i = 0; i < q; i++) {
string s;
cin >> s;
int a = cal(s[0], s[1]);
int b = cal(s[2], s[3]);
int c = cal(s[4], s[5]);
// cout << a << ' ' << b << ' ' << c << endl;
qry[a].emplace_back(b, c, i);
}
priority_queue<Info> Q;
for (int i = 1; i <= n; i++) {
vector<int> d(n + 1, -1);
Q.emplace(i, 0);
while (!Q.empty()) {
auto [u, mx] = Q.top();
Q.pop();
if (d[u] != -1) continue;
d[u] = mx;
for (auto [v, w] : adj[u]) {
Q.emplace(v, max(mx, w));
}
}
for (auto [b, c, idx] : qry[i]) {
if (d[b] != -1 && d[b] <= c) {
ans[idx] = 1;
}
}
}
for (int i = 0; i < q; i++) {
cout << ans[i];
}
cout << "\n";
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 3572kb
input:
2 5 2 4 0305040201 0404040404 030300 020500 050102 020501 6 2 4 030603010601 010203060504 030202 060402 050602 060401
output:
1011 0100
result:
ok 2 lines
Test #2:
score: 0
Accepted
time: 0ms
memory: 3648kb
input:
1 3 3 6 020202 030301 030201 020102 030203 010201 010303 020303 010202
output:
010101
result:
ok single line: '010101'
Test #3:
score: 0
Accepted
time: 68ms
memory: 4080kb
input:
200 10 10 5000 01060:04020305080709 0103070:060204050908 09070503080401060:02 050308010204090:0607 03010502040607080:09 03080109020504060:07 06050:09040302080107 07080305010409060:02 030809010:0204060507 0:060908070201050304 060700 090:03 09080: 070405 010703 0:0100 080601 030600 070206 0:0:09 08040...
output:
011110001101101111111111111111111101111111110111011110110110111011010111111111111111111101111111111110111111110111111111111101111111111110111111111111111111110001100111111111111111111111111011101111111111111111111111111111111111111111011011110100111110111111110111111100111111101110111111111101111110...
result:
ok 200 lines
Test #4:
score: 0
Accepted
time: 93ms
memory: 24700kb
input:
1 2000 1 1000000 M=:]A@8UAY7W2JJ4KEHIA[HSCQ1ENSC`JXR;F3PJ:_@41P9Z=9HR8P<<:DUXRR9^WOQFL?NZP6S@=J0^WE32=6;\U0?88]Q_RNPUMT6YU<4<S]H?:7OCQYOT4YAV1^764ENWSDBED>M7A:BI>KSIR48JQ9B=N\5T3N4A2aF0@>3TI81<G7;YE>W`NMP<:IT4CI3D0=GZC3I\CLQJQBA9BDIS9SAM55KaVA<Z@D=>:Y?CQHUQ5U3a6UVI8OKX9_FAF^7=5M85;<0;8YPAM<7Z7PP7A=N...
output:
000101000101100010001000000010010110000001000001001100000000010000100001000000101100000000010000001000000001110000010110100000111100100000001000000000011001010001000001001000100000000100011001100110010000010000101100000011111000000010000101010010000000010101000001010111100000100000000000000101000100...
result:
ok single line: '000101000101100010001000000010...0101000101000000000010101001000'
Test #5:
score: -100
Time Limit Exceeded
input:
1 2000 2000 1000000 FVAaH7GRPO;_11Da5J18@3SMG==\G8E8S^6:M4L0JH>MN5IXT>2<7WJ3U8LVJS=;;3F13>0D0>VOIIU@EPHG6ABL6;K?T1PXDERLG07]5C9^GDKG<SBMIW;`4W8P3=469TIPKH0O34523_J5C2MJ17D25Z@=K8H@M>WK<CMK7EO]BPD7B6AW741J5YIHIa1:ERSG>L3N2^F3<4F`DLE@2AA5=8GZK6:192FB736[WMV7:^DA2C:<LK040VJBM3M]WXU50`407TR_?PLF@5VL7OSL...